A particle of mass 0.5kg travels …

Body travels in straight line with velocity
v = ax^{3/2}
where a = 5 m-½/s

velocity at x = 0
u { initial velocity } = a(0)^{3/2} = 0

velocity at x = 2m
V { final velocity } = a(2)^{3/2} = 5× (8)½ = 10√2 m/s

we know,
according to Work-energy theorem

workdone = change in kinetic energy
workdone = 1/2 m( Vf² - Vi²)
= 1/2 × 0.5 × { 200 - 0}
=50 Joule

hence, workdone by net Force during 0 ≤ x ≤ 2 is 50 joule

Given,

Mass of the body, m = 0.5 kg
 

Velocity of the body is governed by the equation,

                                  v = ax^3/2

Acceleration, a = 5 m ^-1/2 s^-1

Initial velocity, u (at x = 0) = 0 

Final velocity v (at x = 2 m) = 10√2 m/s 

Work done, W = Change in kinetic energy 

                   = () m (v² - u²) 

                   = () × 0.5 [ (10√2)² - 0²]

                   = () × 0.5 × 10 × 10 × 2

                   = 50 J